A091925 Decimal expansion of Pi^3.

3, 1, 0, 0, 6, 2, 7, 6, 6, 8, 0, 2, 9, 9, 8, 2, 0, 1, 7, 5, 4, 7, 6, 3, 1, 5, 0, 6, 7, 1, 0, 1, 3, 9, 5, 2, 0, 2, 2, 2, 5, 2, 8, 8, 5, 6, 5, 8, 8, 5, 1, 0, 7, 6, 9, 4, 1, 4, 4, 5, 3, 8, 1, 0, 3, 8, 0, 6, 3, 9, 4, 9, 1, 7, 4, 6, 5, 7, 0, 6, 0, 3, 7, 5, 6, 6, 7, 0, 1, 0, 3, 2, 6, 0, 2, 8, 8, 6, 1, 9

Offset: 2

Author

Mohammad K. Azarian, Mar 16 2004

Keywords

• easy
• nonn
• cons

Comments

Surface area of the 6-dimensional unit sphere. - Stanislav Sykora, Nov 08 2013

Surface area of a sphere of diameter Pi equals the volume of the circumscribed cube. - Omar E. Pol, Dec 25 2013

Area of a circle of radius Pi. - Omar E. Pol, Jan 31 2016

Examples

			31.00627668029982017547631506710139520222528856588510769414453810380639...
		

Links

• Harry J. Smith, Table of n, a(n) for n = 2..20000
• G. L. Honaker, Jr. and Chris Caldwell, Prime Curios! 31
• Jean-Christophe Pain, Series representations for Pi^3 involving the golden ratio, arXiv:2206.15281 [math.NT], 2022.
• Khodabakhsh Hessami Pilehrood, Tatiana Hessami Pilehrood, Series acceleration formulas for beta values, Discrete Mathematics & Theoretical Computer Science 12:2 (2010), pp. 223-236.
• Stanislav Sykora, Surface Integrals over n-Dimensional Spheres.
• Index entries for transcendental numbers

Crossrefs

Cf. A000796, A002388, A058285 (continued fraction), A019670, A093954, A092731, A092735.

Programs

Magma

R:= RealField(100); (Pi(R))^3; // G. C. Greubel, Mar 09 2018

Mathematica

First@ RealDigits@ N[Pi^3, 120] (* Michael De Vlieger, Jan 31 2016 *)

PARI

default(realprecision, 20080); x=Pi^3/10; for (n=2, 20000, d=floor(x); x=(x-d)*10; write("b091925.txt", n, " ", d)); \\ Harry J. Smith, Jun 22 2009

Formula

Sum_{k >= 0} binomial(2*k,k)/((2*k + 1)^3*16^k) = 7*Pi^3/216. (Kh. Hessami Pilehrood and T. Hessami Pilehrood).

From Peter Bala, Feb 05 2015: (Start)

The integer sequences A(n) := 2^n*(2*n + 1)!^3/n!^2 and B(n) := A(n)*( Sum {k = 0..n} binomial(2*k,k)*1/(2*k + 1)^3*(1/16)^k ) both satisfy the second order recurrence equation u(n) = (160*n^4 + 128*n^3 + 144*n^2 + 2)*u(n-1) - 32*(n - 1)*(2*n - 1)^7*u(n-2). From this observation we can obtain the continued fraction expansion 7/216*Pi^3 = 1 + 2/(432 - 32*3^7/(4162 - 32*2*5^7/(17714 - ... - 32*(n - 1)*(2*n - 1)^7/((160*n^4 + 128*n^3 + 144*n^2 + 2) - ... )))). Cf. A002388, A019670 and A093954. (End)

From Peter Bala, Oct 31 2019: (Start)

Pi^3 = (1/7) * Sum_{n >= 0} (-1)^n*( 1/(n + 1/6)^3 + 1/(n + 5/6)^3 ).

Pi^3 = (1/31) * Sum_{n >= 0} (-1)^n*( 1/(n + 1/10)^3 - 1/(n + 3/10)^3 - 1/(n + 7/10)^3 + 1/(n + 9/10)^3 ). Cf. A019692, A092731 and A092735. (End)

Equals Integral_{x=-oo..oo} x^2/(exp(x/2) + exp(-x/2)) dx. - Amiram Eldar, May 21 2021